Hybrid All-Pay and Winner-Pay Contests Seminar at DICE in D - - PowerPoint PPT Presentation

Hybrid All-Pay and Winner-Pay Contests

Seminar at DICE in D¨ usseldorf, June 5, 2018 Johan N. M. Lagerl¨

• f
• Dept. of Economics, U. of Copenhagen

Email: johan.lagerlof@econ.ku.dk Website: www.johanlagerlof.com June 2, 2018

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 1 / 24

Introduction: What is a hybrid contest? (1/2)

A hybrid contest:

In some economic, social, or political situation, each one of a number of economic agents try to win an indivisible prize. To increase her probability of winning, each contestant makes both all-pay investments and winner-pay investments.

Example: The competitive bidding to host the Olympic games.

All-pay investments: Candidate cities spend money upfront, with the goal of persuading members of the IOC. Winner-pay investments: A city commits to build new stadia and invest in safety arrangements if being awarded the Games.

To fix ideas, consider the following formalization:

Contestant i chooses xi ≥ 0 and yi ≥ 0 to maximize πi = (vi − yi) pi (s1, s2, . . . sn) − xi, subject to si = f (xi, yi).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 2 / 24

Introduction: What is a hybrid contest? (1/2)

A hybrid contest:

In some economic, social, or political situation, each one of a number of economic agents try to win an indivisible prize. To increase her probability of winning, each contestant makes both all-pay investments and winner-pay investments.

Example: The competitive bidding to host the Olympic games.

All-pay investments: Candidate cities spend money upfront, with the goal of persuading members of the IOC. Winner-pay investments: A city commits to build new stadia and invest in safety arrangements if being awarded the Games.

To fix ideas, consider the following formalization:

Contestant i chooses xi ≥ 0 and yi ≥ 0 to maximize πi = (vi − yi) pi (s1, s2, . . . sn) − xi, subject to si = f (xi, yi).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 2 / 24

Introduction: What is a hybrid contest? (1/2)

A hybrid contest:

In some economic, social, or political situation, each one of a number of economic agents try to win an indivisible prize. To increase her probability of winning, each contestant makes both all-pay investments and winner-pay investments.

Example: The competitive bidding to host the Olympic games.

All-pay investments: Candidate cities spend money upfront, with the goal of persuading members of the IOC. Winner-pay investments: A city commits to build new stadia and invest in safety arrangements if being awarded the Games.

To fix ideas, consider the following formalization:

Contestant i chooses xi ≥ 0 and yi ≥ 0 to maximize πi = (vi − yi) pi (s1, s2, . . . sn) − xi, subject to si = f (xi, yi).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 2 / 24

Introduction: Other examples (2/2)

Competition for a government contract or grant:

All-pay investments: Time/effort spent on preparing proposal. Winner-pay investments: Commit to ambitious customer service.

A political election:

All-pay investments: Campaign expenditures. Winner-pay investments: Electoral promises (costly if they deviate from the politician’s own ideal policy).

Rent seeking to win monopoly rights of a regulated market:

All-pay investments: Ex ante bribes (how Tullock modeled it). Winner-pay investments: Conditional bribes.

Tullock’s motivation:

Empirical studies in the 1950s: DWL appears to be tiny. Tullock: Maybe a part of profits adds to the cost of monopoly.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 3 / 24

Literature Review (1/2)

Two earlier papers that model a hybrid contest:

Haan and Schonbeek (2003).

They assume Cobb-Douglas—which here is quite restrictive.

Melkoyan (2013).

CES but with σ ≥ 1. Symmetric model. Hard to check SOC. My analysis: (i) other approach which yields easy-to-check existence condition; (ii) assumes general production function and CSF; (iii) studies both symmetric and asymmetric models.

Other contest models with more than one influence channel:

Sabotage in contests (improve own performance and sabotage the others performance): Konrad (2000), Chen (2003). War and conflict (choice of production and appropriation): Hirschleifer (1991) and Skaperdas and Syroploulos (1997). Multiple all-pay “arms” (maybe with different costs): Arbatskaya and Mialon (2010).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 4 / 24

Literature Review (1/2)

Two earlier papers that model a hybrid contest:

Haan and Schonbeek (2003).

They assume Cobb-Douglas—which here is quite restrictive.

Melkoyan (2013).

CES but with σ ≥ 1. Symmetric model. Hard to check SOC. My analysis: (i) other approach which yields easy-to-check existence condition; (ii) assumes general production function and CSF; (iii) studies both symmetric and asymmetric models.

Other contest models with more than one influence channel:

Sabotage in contests (improve own performance and sabotage the others performance): Konrad (2000), Chen (2003). War and conflict (choice of production and appropriation): Hirschleifer (1991) and Skaperdas and Syroploulos (1997). Multiple all-pay “arms” (maybe with different costs): Arbatskaya and Mialon (2010).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 4 / 24

Literature Review (2/2)

Multidimensional (procurement) auctions:

Che (2003), Branck (1997), Asker and Cantillon (2008).

Firms bid on both price and (many dimensions of) quality. The components of each bid jointly determine a score. Auctioneer chooses bidder with highest score.

Differences:

In their models, not both all-pay and winner-pay ingredients. Not a probabilistic CSF.

Optimal design of a research contest: Che and Gale (2003).

A principal wants to procure an innovation. Fimrs choose both quality of innovation and the prize if winning. Thus, effectively, both all-pay and winner-pay ingredients. Differences: Not a probabilistic CSF (so mixed strategy eq.), linear production function, mechanism design approach.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 5 / 24

Literature Review (2/2)

Multidimensional (procurement) auctions:

Che (2003), Branck (1997), Asker and Cantillon (2008).

Firms bid on both price and (many dimensions of) quality. The components of each bid jointly determine a score. Auctioneer chooses bidder with highest score.

Differences:

In their models, not both all-pay and winner-pay ingredients. Not a probabilistic CSF.

Optimal design of a research contest: Che and Gale (2003).

A principal wants to procure an innovation. Fimrs choose both quality of innovation and the prize if winning. Thus, effectively, both all-pay and winner-pay ingredients. Differences: Not a probabilistic CSF (so mixed strategy eq.), linear production function, mechanism design approach.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 5 / 24

A model of a hybrid contest (1/2)

n ≥ 2 contestants try to win an indivisible prize. Contestant i chooses xi ≥ 0 and yi ≥ 0 to maximize the following expected payoff: πi = (vi − yi) pi (s) − xi, subject to si = f (xi, yi) , where s = (s1, s2, . . . , sn) and si ≥ 0 is contestant i’s score.

vi > 0 is i’s valuation of the prize. pi (s) is i’s prob. of winning (or contest success function, CSF). xi is the all-pay investment: paid whether i wins or not. yi is the winner-pay investment: paid i.f.f. i wins.

It is a one-shot game where the contestants choose their investments (xi, yi) simultaneously with each other.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 6 / 24

A model of a hybrid contest (2/2)

Assumptions about pi(s):

Twice continuously differentiable in its arguments. Strictly increasing and strictly concave in si. Strictly decreasing in sj for all j = i. The contest is won by someone: n

j=1 pj(s) = 1.

Later I assume that pi(s) is homogeneous in s.

Assumptions about f (xi, yi):

Thrice continuously differentiable in its arguments. Strictly increasing in each of its arguments. Strictly quasiconcave. Homogeneous of degree t > 0: ∀k > 0 f (kxi, kyi) = ktf (xi, yi). Inada conditions to rule out xi = 0 or yi = 0.

Examples: pi(s) = wisr

i

n

j=1 wjsr j

, f (xi, yi) =

• αx

σ−1 σ + (1 − α)y σ−1 σ

tσ

σ−1

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 7 / 24

A model of a hybrid contest (2/2)

Assumptions about pi(s):

Twice continuously differentiable in its arguments. Strictly increasing and strictly concave in si. Strictly decreasing in sj for all j = i. The contest is won by someone: n

j=1 pj(s) = 1.

Later I assume that pi(s) is homogeneous in s.

Assumptions about f (xi, yi):

Thrice continuously differentiable in its arguments. Strictly increasing in each of its arguments. Strictly quasiconcave. Homogeneous of degree t > 0: ∀k > 0 f (kxi, kyi) = ktf (xi, yi). Inada conditions to rule out xi = 0 or yi = 0.

Examples: pi(s) = wisr

i

n

j=1 wjsr j

, f (xi, yi) =

• αx

σ−1 σ + (1 − α)y σ−1 σ

tσ

σ−1

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 7 / 24

A model of a hybrid contest (2/2)

Assumptions about pi(s):

Twice continuously differentiable in its arguments. Strictly increasing and strictly concave in si. Strictly decreasing in sj for all j = i. The contest is won by someone: n

j=1 pj(s) = 1.

Later I assume that pi(s) is homogeneous in s.

Assumptions about f (xi, yi):

Thrice continuously differentiable in its arguments. Strictly increasing in each of its arguments. Strictly quasiconcave. Homogeneous of degree t > 0: ∀k > 0 f (kxi, kyi) = ktf (xi, yi). Inada conditions to rule out xi = 0 or yi = 0.

Examples: pi(s) = wisr

i

n

j=1 wjsr j

, f (xi, yi) =

• αx

σ−1 σ + (1 − α)y σ−1 σ

tσ

σ−1

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 7 / 24

Analysis (1/7)

One possible approach:

Plug the production function into the CSF. Take FOCs w.r.t. xi and yi. Used by Haan and Schoonbeek (2003) and Melkoyan (2013), assuming Cobb-Douglas and CES, respectively.

My approach: Solve for contestant i’s best reply in two steps:

1 Compute the conditional factor demands.

That is, derive optimal xi and yi, given s (so also given si).

2 Plug the factor demands into the payoff and then characterize

contestant i’s optimal score si (given s−i).

Important advantage: a single choice variable at 2, so easier to determine what conditions are required for equilibrium existence.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 8 / 24

Analysis (1/7)

One possible approach:

Plug the production function into the CSF. Take FOCs w.r.t. xi and yi. Used by Haan and Schoonbeek (2003) and Melkoyan (2013), assuming Cobb-Douglas and CES, respectively.

My approach: Solve for contestant i’s best reply in two steps:

1 Compute the conditional factor demands.

That is, derive optimal xi and yi, given s (so also given si).

2 Plug the factor demands into the payoff and then characterize

contestant i’s optimal score si (given s−i).

Important advantage: a single choice variable at 2, so easier to determine what conditions are required for equilibrium existence.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 8 / 24

Contestant i solves (for fixed pi): minxi,yi piyi + xi, subject to f (xi, yi) = si. The first-order conditions (λi is the Lagrange multiplier): ∂Li ∂xi = 1 − λif1 (xi, yi) = 0, ∂Li ∂yi = pi − λif2 (xi, yi) = 0. So, by combining the FOCs: 1 pi = f1 (xi, yi) f2 (xi, yi)

def

= g xi yi

• ⇒ xi = yih

1 pi

• ,

where h is the inverse of g (i.e., h

def

= g −1). By plugging back into si = f (xi, yi) and rewriting, we obtain: Yi (si, pi) =

• si

f (h (1/pi) , 1) 1

t

, Xi (si, pi) = Yi (si, pi) h 1 pi

• .

Contestant i’s payoff: πi (s) = pi (s) vi − Ci [si, pi (s)], where Ci [si, pi (s)]

def

= pi (s) Yi [si, pi (s)] + Xi [si, pi (s)] . A Nash equilibrium of the hybrid contest:

A profile s∗ such that πi (s∗) ≥ πi

• si, s∗

−i

• , all i and all si ≥ 0.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 9 / 24

Contestant i solves (for fixed pi): minxi,yi piyi + xi, subject to f (xi, yi) = si. The first-order conditions (λi is the Lagrange multiplier): ∂Li ∂xi = 1 − λif1 (xi, yi) = 0, ∂Li ∂yi = pi − λif2 (xi, yi) = 0. So, by combining the FOCs: 1 pi = f1 (xi, yi) f2 (xi, yi)

def

= g xi yi

• ⇒ xi = yih

1 pi

• ,

where h is the inverse of g (i.e., h

def

= g −1). By plugging back into si = f (xi, yi) and rewriting, we obtain: Yi (si, pi) =

• si

f (h (1/pi) , 1) 1

t

, Xi (si, pi) = Yi (si, pi) h 1 pi

• .

Contestant i’s payoff: πi (s) = pi (s) vi − Ci [si, pi (s)], where Ci [si, pi (s)]

def

= pi (s) Yi [si, pi (s)] + Xi [si, pi (s)] . A Nash equilibrium of the hybrid contest:

A profile s∗ such that πi (s∗) ≥ πi

• si, s∗

−i

• , all i and all si ≥ 0.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 9 / 24

Contestant i solves (for fixed pi): minxi,yi piyi + xi, subject to f (xi, yi) = si. The first-order conditions (λi is the Lagrange multiplier): ∂Li ∂xi = 1 − λif1 (xi, yi) = 0, ∂Li ∂yi = pi − λif2 (xi, yi) = 0. So, by combining the FOCs: 1 pi = f1 (xi, yi) f2 (xi, yi)

def

= g xi yi

• ⇒ xi = yih

1 pi

• ,

where h is the inverse of g (i.e., h

def

= g −1). By plugging back into si = f (xi, yi) and rewriting, we obtain: Yi (si, pi) =

• si

f (h (1/pi) , 1) 1

t

, Xi (si, pi) = Yi (si, pi) h 1 pi

• .

Contestant i’s payoff: πi (s) = pi (s) vi − Ci [si, pi (s)], where Ci [si, pi (s)]

def

= pi (s) Yi [si, pi (s)] + Xi [si, pi (s)] . A Nash equilibrium of the hybrid contest:

A profile s∗ such that πi (s∗) ≥ πi

• si, s∗

−i

• , all i and all si ≥ 0.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 9 / 24

Contestant i solves (for fixed pi): minxi,yi piyi + xi, subject to f (xi, yi) = si. The first-order conditions (λi is the Lagrange multiplier): ∂Li ∂xi = 1 − λif1 (xi, yi) = 0, ∂Li ∂yi = pi − λif2 (xi, yi) = 0. So, by combining the FOCs: 1 pi = f1 (xi, yi) f2 (xi, yi)

def

= g xi yi

• ⇒ xi = yih

1 pi

• ,

where h is the inverse of g (i.e., h

def

= g −1). By plugging back into si = f (xi, yi) and rewriting, we obtain: Yi (si, pi) =

• si

f (h (1/pi) , 1) 1

t

, Xi (si, pi) = Yi (si, pi) h 1 pi

• .

Contestant i’s payoff: πi (s) = pi (s) vi − Ci [si, pi (s)], where Ci [si, pi (s)]

def

= pi (s) Yi [si, pi (s)] + Xi [si, pi (s)] . A Nash equilibrium of the hybrid contest:

A profile s∗ such that πi (s∗) ≥ πi

• si, s∗

−i

• , all i and all si ≥ 0.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 9 / 24

Contestant i solves (for fixed pi): minxi,yi piyi + xi, subject to f (xi, yi) = si. The first-order conditions (λi is the Lagrange multiplier): ∂Li ∂xi = 1 − λif1 (xi, yi) = 0, ∂Li ∂yi = pi − λif2 (xi, yi) = 0. So, by combining the FOCs: 1 pi = f1 (xi, yi) f2 (xi, yi)

def

= g xi yi

• ⇒ xi = yih

1 pi

• ,

where h is the inverse of g (i.e., h

def

= g −1). By plugging back into si = f (xi, yi) and rewriting, we obtain: Yi (si, pi) =

• si

f (h (1/pi) , 1) 1

t

, Xi (si, pi) = Yi (si, pi) h 1 pi

• .

Contestant i’s payoff: πi (s) = pi (s) vi − Ci [si, pi (s)], where Ci [si, pi (s)]

def

= pi (s) Yi [si, pi (s)] + Xi [si, pi (s)] . A Nash equilibrium of the hybrid contest:

A profile s∗ such that πi (s∗) ≥ πi

• si, s∗

−i

• , all i and all si ≥ 0.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 9 / 24

Contestant i solves (for fixed pi): minxi,yi piyi + xi, subject to f (xi, yi) = si. The first-order conditions (λi is the Lagrange multiplier): ∂Li ∂xi = 1 − λif1 (xi, yi) = 0, ∂Li ∂yi = pi − λif2 (xi, yi) = 0. So, by combining the FOCs: 1 pi = f1 (xi, yi) f2 (xi, yi)

def

= g xi yi

• ⇒ xi = yih

1 pi

• ,

where h is the inverse of g (i.e., h

def

= g −1). By plugging back into si = f (xi, yi) and rewriting, we obtain: Yi (si, pi) =

• si

f (h (1/pi) , 1) 1

t

, Xi (si, pi) = Yi (si, pi) h 1 pi

• .

Contestant i’s payoff: πi (s) = pi (s) vi − Ci [si, pi (s)], where Ci [si, pi (s)]

def

= pi (s) Yi [si, pi (s)] + Xi [si, pi (s)] . A Nash equilibrium of the hybrid contest:

A profile s∗ such that πi (s∗) ≥ πi

• si, s∗

−i

• , all i and all si ≥ 0.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 9 / 24

Analysis (3/7)

The cost-minimization problem and the h function

xi yi slope = − 1

pi

slope = −g

• xi

yi

• si = f (xi, yi)

X Y

(a) Cost minimization.

xi yi

m g

• xi

yi

• 45◦

(b) Graph of the g function.

m

xi yi

h (m) 45◦

(c) Graph of the h function.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 10 / 24

Analysis (4/7)

Equilibrium existence Define the following elasticities: The elasticity of output w.r.t. xi: η

• 1

pi

• def

=

f1

• h
• 1

pi

• ,1
• h
• 1

pi

• f
• h
• 1

pi

• ,1
• .

The elasticity of substitution: σ

• 1

pi

• def

= −

h′

1 pi

• 1

pi

h

• 1

pi

.

The elasticity of the win probability w.r.t. si: εi (s)

def

= ∂pi

∂si si pi .

We have that η ∈ (0, t), σ > 0, and εi ∈ (0, 1). Assumption 1. The production function and the CSF satisfy:

(i) t ≤ 1 and εi (s) η

• 1

pi

• σ
• 1

pi

• ≤ 2

(for all pi and s);

Proposition 1. Suppose Assumption 1 is satisfied. Then there exists a pure strategy Nash equilibrium of the hybrid contest.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 11 / 24

Analysis (4/7)

Equilibrium existence Define the following elasticities: The elasticity of output w.r.t. xi: η

• 1

pi

• def

=

f1

• h
• 1

pi

• ,1
• h
• 1

pi

• f
• h
• 1

pi

• ,1
• .

The elasticity of substitution: σ

• 1

pi

• def

= −

h′

1 pi

• 1

pi

h

• 1

pi

.

The elasticity of the win probability w.r.t. si: εi (s)

def

= ∂pi

∂si si pi .

We have that η ∈ (0, t), σ > 0, and εi ∈ (0, 1). Assumption 1. The production function and the CSF satisfy:

(i) t ≤ 1 and εi (s) η

• 1

pi

• σ
• 1

pi

• ≤ 2

(for all pi and s);

Proposition 1. Suppose Assumption 1 is satisfied. Then there exists a pure strategy Nash equilibrium of the hybrid contest.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 11 / 24

Analysis (4/7)

Equilibrium existence Define the following elasticities: The elasticity of output w.r.t. xi: η

• 1

pi

• def

=

f1

• h
• 1

pi

• ,1
• h
• 1

pi

• f
• h
• 1

pi

• ,1
• .

The elasticity of substitution: σ

• 1

pi

• def

= −

h′

1 pi

• 1

pi

h

• 1

pi

.

The elasticity of the win probability w.r.t. si: εi (s)

def

= ∂pi

∂si si pi .

We have that η ∈ (0, t), σ > 0, and εi ∈ (0, 1). Assumption 1. The production function and the CSF satisfy:

(i) t ≤ 1 and εi (s) η

• 1

pi

• σ
• 1

pi

• ≤ 2

(for all pi and s);

Proposition 1. Suppose Assumption 1 is satisfied. Then there exists a pure strategy Nash equilibrium of the hybrid contest.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 11 / 24

Analysis (4/7)

Equilibrium existence Define the following elasticities: The elasticity of output w.r.t. xi: η

• 1

pi

• def

=

f1

• h
• 1

pi

• ,1
• h
• 1

pi

• f
• h
• 1

pi

• ,1
• .

The elasticity of substitution: σ

• 1

pi

• def

= −

h′

1 pi

• 1

pi

h

• 1

pi

.

The elasticity of the win probability w.r.t. si: εi (s)

def

= ∂pi

∂si si pi .

We have that η ∈ (0, t), σ > 0, and εi ∈ (0, 1). Assumption 1. The production function and the CSF satisfy:

(i) t ≤ 1 and εi (s) η

• 1

pi

• σ
• 1

pi

• ≤ 2

(for all pi and s);

Proposition 1. Suppose Assumption 1 is satisfied. Then there exists a pure strategy Nash equilibrium of the hybrid contest.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 11 / 24

Assume a CES production function, t = 1, r ≤ 1, and pi(s) = wisr

i

n

j=1 wjsr j

and pi(0, ∙ ∙ ∙ , 0) = wi n

j=1 wj

. σ α

2 r 4 r

σ∗

15 r 20 r 1 4

α∗

3 4

1 Θ(σ, r)

def

= (

2 rσ−2) 1 σ

1+(

2 rσ−2) 1 σ

Assumption 1 satisfied

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 12 / 24

Analysis (6/7)

To check the SOC with Melkoyan’s analytical approach is cumbersome and in the end he relies on numerical simulations:

[. . . ] one can demonstrate, after a series of tedious algebraic manipulations, that a player’s payoff function is locally concave at the symmetric equilibrium candidate in (7) if and only if [large mathematical expression]. [. . . ] Numerical simulations indicate that this inequality is violated only for extreme values of the parameters [. . . ]. In addition to verifying the local second-order conditions, I have used numerical simulations to verify that the global second-order conditions are satisfied under a wide range of scenarios.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 13 / 24

Characterization of equilibrium Recall: πi (s) = pi (s) vi − Ci [si, pi (s)]. The FOC (with an equality if si > 0): ∂πi (s) ∂si = ∂pi (s) ∂si vi − C1 (si, pi) − C2 (si, pi) ∂pi (s) ∂si ≤ 0. Use Shephard’s lemma, C2 (si, pi) = Yi [si, pi (s)]: [vi − Yi (si, pi (s))] ∂pi (s) ∂si ≤ C1 (si, pi) , (1) with an equality if si > 0. Proposition 2. Suppose Assumption 1 is satisfied. Then s∗ = (s∗

1, . . . , s∗ n) is a pure strategy Nash equilibrium of the

hybrid contest if and only if condition (1) holds, with equality if s∗

i > 0, for each contestant i.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 14 / 24

Characterization of equilibrium Recall: πi (s) = pi (s) vi − Ci [si, pi (s)]. The FOC (with an equality if si > 0): ∂πi (s) ∂si = ∂pi (s) ∂si vi − C1 (si, pi) − C2 (si, pi) ∂pi (s) ∂si ≤ 0. Use Shephard’s lemma, C2 (si, pi) = Yi [si, pi (s)]: [vi − Yi (si, pi (s))] ∂pi (s) ∂si ≤ C1 (si, pi) , (1) with an equality if si > 0. Proposition 2. Suppose Assumption 1 is satisfied. Then s∗ = (s∗

1, . . . , s∗ n) is a pure strategy Nash equilibrium of the

hybrid contest if and only if condition (1) holds, with equality if s∗

i > 0, for each contestant i.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 14 / 24

Characterization of equilibrium Recall: πi (s) = pi (s) vi − Ci [si, pi (s)]. The FOC (with an equality if si > 0): ∂πi (s) ∂si = ∂pi (s) ∂si vi − C1 (si, pi) − C2 (si, pi) ∂pi (s) ∂si ≤ 0. Use Shephard’s lemma, C2 (si, pi) = Yi [si, pi (s)]: [vi − Yi (si, pi (s))] ∂pi (s) ∂si ≤ C1 (si, pi) , (1) with an equality if si > 0. Proposition 2. Suppose Assumption 1 is satisfied. Then s∗ = (s∗

1, . . . , s∗ n) is a pure strategy Nash equilibrium of the

hybrid contest if and only if condition (1) holds, with equality if s∗

i > 0, for each contestant i.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 14 / 24

Characterization of equilibrium Recall: πi (s) = pi (s) vi − Ci [si, pi (s)]. The FOC (with an equality if si > 0): ∂πi (s) ∂si = ∂pi (s) ∂si vi − C1 (si, pi) − C2 (si, pi) ∂pi (s) ∂si ≤ 0. Use Shephard’s lemma, C2 (si, pi) = Yi [si, pi (s)]: [vi − Yi (si, pi (s))] ∂pi (s) ∂si ≤ C1 (si, pi) , (1) with an equality if si > 0. Proposition 2. Suppose Assumption 1 is satisfied. Then s∗ = (s∗

1, . . . , s∗ n) is a pure strategy Nash equilibrium of the

hybrid contest if and only if condition (1) holds, with equality if s∗

i > 0, for each contestant i.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 14 / 24

Characterization of equilibrium Recall: πi (s) = pi (s) vi − Ci [si, pi (s)]. The FOC (with an equality if si > 0): ∂πi (s) ∂si = ∂pi (s) ∂si vi − C1 (si, pi) − C2 (si, pi) ∂pi (s) ∂si ≤ 0. Use Shephard’s lemma, C2 (si, pi) = Yi [si, pi (s)]: [vi − Yi (si, pi (s))] ∂pi (s) ∂si ≤ C1 (si, pi) , (1) with an equality if si > 0. Proposition 2. Suppose Assumption 1 is satisfied. Then s∗ = (s∗

1, . . . , s∗ n) is a pure strategy Nash equilibrium of the

hybrid contest if and only if condition (1) holds, with equality if s∗

i > 0, for each contestant i.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 14 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

A Symmetric Hybrid Contest (1/4)

Assumption 2. The CSF is symmetric and homogeneous of degree 0. Note that, thanks to Assumption 2: ∂pi(s, s, . . . , s) ∂si = ε(n) ns , where ε(n)

def

= εi (1, 1, . . . , 1) . Use this in the FOC and impose symmetry: (v − y ∗) ε(n) ns∗ = C1

• s∗, 1

n

• = 1

ts∗C

• s∗, 1

n

• = 1

ts∗ y ∗ n + x∗

• ⇔ (v − y ∗) t

ε(n) = y ∗ + nx∗. And from before, x∗ = h(n)y ∗. The last equalities are linear in x∗ and y ∗, so easy to solve. Proposition 3. Within the family of sym. eq., there is a unique pure strategy equilibrium: s∗ = f [h(n), 1] (y ∗)t, x∗ = h(n)y ∗, and y ∗ = t ε(n)v 1 + nh(n) + t ε(n).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 15 / 24

Proposition 4. Effect of more contestants on x∗ and y ∗: ∂x∗ ∂n < 0 ⇔ σ(n) > − n (n − 2) h(n) − 1 (n − 1) [1 + t ε(n)], ∂y ∗ ∂n > 0 ⇔ σ(n) > n(n − 2)h(n) − 1 (n − 1)nh(n) ; and if σ(n) ≥ 1, then necessarily ∂x∗

∂n < 0 and ∂y∗ ∂n > 0.

In order to understand the above:

More contestants means a lower probability of winning. This lowers the relative cost of investing in yi. So whenever σ(n) is sufficiently large, ∂y∗

∂n > 0 and ∂x∗ ∂n < 0.

But if σ(n) small, the derivatives must have the same sign. For: ∂y∗ ∂n n y∗ = σ(n) + ∂x∗ ∂n n x∗ (follows from x∗ = h(n)y∗). As σ(n) → 0, the production function requires xi and yi to be used in fixed proportions (a Leontief production technology).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 16 / 24

The total amount of equilibrium expenditures in the symmetric hybrid model is defined as RH def = nC

• s∗, 1

n

• .

The corresponding amount in the all-pay contest: RA = t ε(n)v. Proposition 5, part (a). In the symmetric model: RH = (1 − y ∗ v )RA =

• 1

v [1 + nh(n)] + 1 RA −1 . In particular, for any finite n, we have RH < RA.

The payoff suggests the intuition: πi = (vi − yi) pi (s) − xi.

Proposition 5, part (b). In the symmetric model, suppose pi(s) = φ(si)/ n

j=1 φ(sj), where φ is a strictly increasing and

concave function satisfying φ(0) = 0.

Then RH is weakly increasing in n if and only if: (i) σ (n) ≤ 1 + 4n tr (n − 1)2 ; (2)

• r (ii) inequality (2) is violated and h (n) /

∈ (ΞL, ΞH). See figure!

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 17 / 24

The total amount of equilibrium expenditures in the symmetric hybrid model is defined as RH def = nC

• s∗, 1

n

• .

The corresponding amount in the all-pay contest: RA = t ε(n)v. Proposition 5, part (a). In the symmetric model: RH = (1 − y ∗ v )RA =

• 1

v [1 + nh(n)] + 1 RA −1 . In particular, for any finite n, we have RH < RA.

The payoff suggests the intuition: πi = (vi − yi) pi (s) − xi.

Proposition 5, part (b). In the symmetric model, suppose pi(s) = φ(si)/ n

j=1 φ(sj), where φ is a strictly increasing and

concave function satisfying φ(0) = 0.

Then RH is weakly increasing in n if and only if: (i) σ (n) ≤ 1 + 4n tr (n − 1)2 ; (2)

• r (ii) inequality (2) is violated and h (n) /

∈ (ΞL, ΞH). See figure!

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 17 / 24

A Symmetric Hybrid Contest (4/4)

Illustration of result (b) Assume CES, t = 1, and n = 10. σ α 2 4 1.494

1 4 1 2 3 4

1 Assumption 1 satisfied R

H

d e c r e a s i n g i n n a t n = 1

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 18 / 24

Asymmetric Hybrid Contests (1/2)

I assume n = 2 and I study three models:

The CSF is biased in favor of one contestant. One contestant has a higher valuation than the other. I also endogenize the degree of bias.

Assumption 3. The CSF is given by pi(s) = wisr

i

w1sr

1 + w2sr 2

.

The following three equations define equilibrium values of p∗

1, y ∗ 1 , and y ∗ 2 :

y ∗

i =

rtp∗

i (1 − p∗ i )vi

rtp∗

i (1 − p∗ i ) + p∗ i + h

• 1

p∗

i

, for i = 1, 2, and Υ(p∗

1) = 0, where

Υ(p1) def =

w2vrt

2

w1vrt

1 p1f

• h
• 1

1−p1

• , 1

r

• rtp1(1 − p1) + 1 − p1 + h
• 1

1−p1

rt − (1 − p1)f

• h
• 1

p1

• , 1

r

• rtp1(1 − p1) + p1 + h
• 1

p1

rt . The equilibrium is unique if rη

• 1

pi

• σ
• 1

pi

• ≤ 1.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 19 / 24

Asymmetric Hybrid Contests (1/2)

I assume n = 2 and I study three models:

The CSF is biased in favor of one contestant. One contestant has a higher valuation than the other. I also endogenize the degree of bias.

Assumption 3. The CSF is given by pi(s) = wisr

i

w1sr

1 + w2sr 2

.

The following three equations define equilibrium values of p∗

1, y ∗ 1 , and y ∗ 2 :

y ∗

i =

rtp∗

i (1 − p∗ i )vi

rtp∗

i (1 − p∗ i ) + p∗ i + h

• 1

p∗

i

, for i = 1, 2, and Υ(p∗

1) = 0, where

Υ(p1) def =

w2vrt

2

w1vrt

1 p1f

• h
• 1

1−p1

• , 1

r

• rtp1(1 − p1) + 1 − p1 + h
• 1

1−p1

rt − (1 − p1)f

• h
• 1

p1

• , 1

r

• rtp1(1 − p1) + p1 + h
• 1

p1

rt . The equilibrium is unique if rη

• 1

pi

• σ
• 1

pi

• ≤ 1.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 19 / 24

Asymmetric Hybrid Contests (1/2)

I assume n = 2 and I study three models:

The CSF is biased in favor of one contestant. One contestant has a higher valuation than the other. I also endogenize the degree of bias.

Assumption 3. The CSF is given by pi(s) = wisr

i

w1sr

1 + w2sr 2

.

The following three equations define equilibrium values of p∗

1, y ∗ 1 , and y ∗ 2 :

y ∗

i =

rtp∗

i (1 − p∗ i )vi

rtp∗

i (1 − p∗ i ) + p∗ i + h

• 1

p∗

i

, for i = 1, 2, and Υ(p∗

1) = 0, where

Υ(p1) def =

w2vrt

2

w1vrt

1 p1f

• h
• 1

1−p1

• , 1

r

• rtp1(1 − p1) + 1 − p1 + h
• 1

1−p1

rt − (1 − p1)f

• h
• 1

p1

• , 1

r

• rtp1(1 − p1) + p1 + h
• 1

p1

rt . The equilibrium is unique if rη

• 1

pi

• σ
• 1

pi

• ≤ 1.
• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 19 / 24

Asymmetric Hybrid Contests (2/4)

A Biased decision process (w1 = w2 but v1 = v2) Among the results:

(a) p∗

1 > p∗ 2 ⇔ y∗ 1 < y∗ 2 ⇔ C(s∗ 1, p∗ 1) > C(s∗ 2, p∗ 2).

(b) Evaluated at symmetry (w1 = w2): ∂p∗

1

∂w1 > 0,

∂y ∗

1

∂w1 < 0, ∂y ∗

2

∂w1 > 0, ∂x∗

1

∂w1 > 0 ⇔ ∂x∗

2

∂w1 < 0 ⇔ σ(2) > 2 2 + rt .

Different valuations (v1 = v2 but w1 = w2) Among the results:

(a) p∗

1 > p∗ 2 ⇔ y∗

1

v1 < y∗

2

v2 .

(b) v1 − y∗

1 > v2 − y∗ 2 ⇔ C(s∗ 1, p∗ 1) > C(s∗ 2, p∗ 2).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 20 / 24

Asymmetric Hybrid Contests (2/4)

A Biased decision process (w1 = w2 but v1 = v2) Among the results:

(a) p∗

1 > p∗ 2 ⇔ y∗ 1 < y∗ 2 ⇔ C(s∗ 1, p∗ 1) > C(s∗ 2, p∗ 2).

(b) Evaluated at symmetry (w1 = w2): ∂p∗

1

∂w1 > 0,

∂y ∗

1

∂w1 < 0, ∂y ∗

2

∂w1 > 0, ∂x∗

1

∂w1 > 0 ⇔ ∂x∗

2

∂w1 < 0 ⇔ σ(2) > 2 2 + rt .

Different valuations (v1 = v2 but w1 = w2) Among the results:

(a) p∗

1 > p∗ 2 ⇔ y∗

1

v1 < y∗

2

v2 .

(b) v1 − y∗

1 > v2 − y∗ 2 ⇔ C(s∗ 1, p∗ 1) > C(s∗ 2, p∗ 2).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 20 / 24

An Endogenous Bias (w1 chosen, but v1 ≥ v2 and w2 fixed) Timing of events in the game:

1 A principal chooses w1 to maximize RH = C(s∗ 1, p∗ 1) + C(s∗ 2, p∗ 2). 2 w1 becomes common knowledge and the contestants interact as

in the previous analysis.

Assumption 3. The production function is of Cobb-Douglas form: f (xi, yi) = xα

i y β i , for α > 0 and β > 0.

Results: The equilibrium values of p1 and w1 satisfy:

If v1 = v2, then p1 = 1

2 and

w1 = w2. If v1 > v2, then p1 > 1

2.

If v1 > v2, then w1 < w2 at least if |v1 − v2| is very small or big.

My intuition for results:

Contestant 1 is more valuable as a contributor (as v1 > v2). Hence, she should be encouraged to use x1, as all-pay investments are more conducive to large expenditures. This is achieved by making winner-pay inv. costly: p1 > 1

2.

To generate p1 > 1

2, v1 > v2 is more than enough, so bias can

be in favor of Contestant 2.

Might not be robust.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 21 / 24

An Endogenous Bias (w1 chosen, but v1 ≥ v2 and w2 fixed) Timing of events in the game:

1 A principal chooses w1 to maximize RH = C(s∗ 1, p∗ 1) + C(s∗ 2, p∗ 2). 2 w1 becomes common knowledge and the contestants interact as

in the previous analysis.

Assumption 3. The production function is of Cobb-Douglas form: f (xi, yi) = xα

i y β i , for α > 0 and β > 0.

Results: The equilibrium values of p1 and w1 satisfy:

If v1 = v2, then p1 = 1

2 and

w1 = w2. If v1 > v2, then p1 > 1

2.

If v1 > v2, then w1 < w2 at least if |v1 − v2| is very small or big.

My intuition for results:

Contestant 1 is more valuable as a contributor (as v1 > v2). Hence, she should be encouraged to use x1, as all-pay investments are more conducive to large expenditures. This is achieved by making winner-pay inv. costly: p1 > 1

2.

To generate p1 > 1

2, v1 > v2 is more than enough, so bias can

be in favor of Contestant 2.

Might not be robust.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 21 / 24

Numerical example (t = r = v2 = w2 = 1) Plot of plot p1 and w1 against v1 for three different values of α: 0.9 (the blue, dotted curve), 0.5 (the green, dashed curve), and 0.1 (the red, solid curve). v1

• p1

0012345678910 .50 .55

(a) The high-valuation contestant’s probability of winning.

v1

• w1

012345678910 .25 .50 .75 1

(b) The weight in the CSF that is assigned to the high-valuation contestant’s score.

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 22 / 24

Main results and contributions: (1/1)

1 The analytical approach (borrowing from producer theory):

→ Generality, tractability, and an existence condition.

2 A larger n leads to substitution away from all-pay investments.

But only if the elasticity of substitution is large enough.

3 Total expenditures always lower in hybrid contest than in all-pay. 4 Total exp’tures can be decreasing in n (also shown by Melkoyan). 5 Asym. contests (in terms of valuations and bias): Sharp

predictions about relative size of investm’s and of expenditures.

6 Endogenous bias: High-valuation contestant more likely to win

but the bias is against her (the latter might not be robust).

• J. Lagerl¨
• f (U of Copenhagen)

Hybrid All-Pay and Winner-Pay Contests June 5, 2018 23 / 24

Possible avenues for future work (1/1)

1 Sequential moves: first (x1, y1), then (x2, y2). 2 Applying the producer theory approach to other contest models

with multiple influence channels.

3 Experimental testing.

Relatively sharp predictions. But risk neutrality might be an issue?

4 Further work on asymmetric contests.

More than two contestants. Can a contestant be hurt by a bias in favor of her? Can a contestant benefit from an increase in rival’s valuation?

5 Contest design in broader settings.

• J. Lagerl¨
• f (U of Copenhagen)

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